Question: Find $\lim_{x\to\infty}\dfrac{x^2+1}{\sin(x)}$. Choose 1 answer: Choose 1 answer: (Choice A) A $0$ (Choice B) B $1$ (Choice C) C $2$ (Choice D) D The limit doesn't exist
Solution: When dealing with limits that include $\sin(x)$, it's important to remember that $\lim_{x\to\infty}\sin(x)$ doesn't exist, as $\sin(x)$ keeps oscillating between $-1$ and $1$ forever. ${2}$ ${4}$ ${6}$ ${8}$ ${\llap{-}4}$ ${\llap{-}6}$ ${\llap{-}8}$ ${2}$ $y$ $x$ $y=\sin(x)$ This doesn't necessarily mean that our limit doesn't exist. Think what happens to $\dfrac{x^2+1}{\sin(x)}$ as $x$ increases towards positive infinity. While $x^2+1$ keeps growing boundlessly, $\sin(x)$ oscillates from $-1$, to $0$, to $1$, to $0$, to $-1$ again. The result is a graph that goes up and down forever, with vertical asymptotes every now and then. ${10}$ ${20}$ ${30}$ ${40}$ ${\llap{-}10}$ ${\llap{-}20}$ ${\llap{-}30}$ ${\llap{-}40}$ ${2000}$ ${4000}$ ${\llap{-}2000}$ ${\llap{-}4000}$ $y$ $x$ $y=\dfrac{x^2+1}{\sin(x)}$ This limit doesn't approach any specific value as $x$ increases towards infinity. In conclusion, $\lim_{x\to\infty}\dfrac{x^2+1}{\sin(x)}$ doesn't exist.